Book I.

POSTULATES.

L

I.
ET it be granted that a straight line may be drawn from
any one point to any other point.

II.
That a terminated straight line may be produced to any length
in a straight line.

III.
And that a circle may be described from any centre, at any

distance from that centre.'

A X I O M S.

T

I.
THINGS which are equal to the same, are equal to one an-
other.

II.
If equals be added to equals, the wholes are equal.

III.
If equals be taken from equals, the remainders are equal.

IV.
If equals be added to unequals, the wholes are unequal.

V.
If equals be taken from unequals, the remainders are unequal.

VI.
Things which are double of the fame, are equal to one another,

VII.
Things which are halves of the fame, are equal to one another,

VIII.
Magnitudes which coincide with one another, that is, which
exactly fill the same space, are equal to one another.

IX. The whole is greater than its part.

X. Straight lines which do not coincide, cannot meet one another in See N. more than one point.

XI.
All right angles are equal to one another.

XII.
“ If a straight line meets two straight lines, so as to make the

“ two interior angles on the same side of it taken together, “ less than two right angles, these straight lines, being con

tinually produced, shall at length meet upon that side on “ which are the angles which are less than two right angles. " See the notes on Prop. 29. of Book 1.".

PROPO.

Book I.

PROPOSITION I. PROBLEM.
To describe an equilateral triangle upon a given

finite itraight line.
Let AB be the given straight line; it is required to describe
an equilateral triangle upon it.

. From the centre A, at the di

с.
a 3. Postu. flance AB, describe a the circle
lale.

BCD; and from the centre B, at
the distance BA, describe the
circle ACE ; and from the point

B E C, in which the circles cut one b 1. Poft. another, draw the straight lines b

CA, CB to the points A, B; ABC
shall be an equilateral triangle.

Because the point A is the centre of the circle BCD, AC is cisth De- equal c to AB; and because the point B is the centre of the finition.

circle ACE, BC is equal to BA: But it has been proved, that CA is equal to AB; therefore CA, CB are each of them equal

to AB; but things which are equal to the same, are equal to one df Arj. another d; therefore CA is equal to CB; wherefore CA, AB,

BC, are equal to one another; and the triangle ABC is there-
fore equilateral, and it is described upon the given straight line
AB. Which was required to be done.

PROP. II. PROB.
ROM a given point to draw a straight line equal to

a given itraight line.
Let A be the given point, and BC the given straight line ; it is

required to draw from the point A a straight line equal to BC. a . Poft.

From the point A to B draw a the straight line AB; and upon it

K describe b the equilateral triangle C 2. Poft. DAB, and produce the straight

C

H
lines DA, DB, to E and F; from
the centre B, at the distance BC,

D А
d 3. Poft. describe d the circle CGH; and С
from the centre D, at the distance

B
DG, describe the circle GKL. AL

G
shall be equal to BC,

F

b 1.1.

F

Because

e

Because the point B is the centre of the circle CGH, BC is Book I. equal e to BG; and because D is the centre of the circle GKL, Word DL is equal to DG, and DA, DB, parts of them, are equal ; e 15. Def. therefore the remainder AL is equal to the remainder f BG:f 3. Ax. But it has been shewn, that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the fame, are equal to one another; therefore the straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the givea straight line BC. Which was to be done.

F

E

a 2. I.

PROP. III. PROB.
TROM the greater of two given straight lines to

cut off a part equal to the less.
Let AB and C be the two given
straight lines, whereof AB is the greater:
It is required to cut off from AB, the
greater, a part equal to C, the lefs.

A From the point A draw a the straight

B line AD equal to C; and from the centre A, at the distance AD, describe b

b 3. Poft, the circle DEF: And because A is the centre of the circle DEF, AE is equal to AD; but the straight line C is likewise equal to AD; whence AE and C are each of them equal to AD; wherefore the ftraight line AE is equal to c C, and from AB, C 1. Ax. the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to be done.

PROP. IV. THEOREM.
F two triangles have two sides of the one equal

to two sides of the other, each to each ; and have likewise the angles contained by those fides equal to one another; they shall likewise have their bases, or third fides, equal; and the two triangles shall be equal; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite.

Let ABC, DEF be two triangles, which have the two sides AB, AC equal to the two fides DE, DF, each to each, viz.

AB

IF

Book I. AB to DE, and AC to

A
DF; and the angle BAC
equal to the angle EDF;
the base BC shall be equal
to the base EF; and the
triangle ABC to the tri-
angle DEF; and the other
angles, to which the equal

B

C E
sides are opposite, shall be
equal each to each, viz.
the angle ABC to the angle
DEF, and the angle ACB
to DFE.

For, if the triangle ABC be applied to DEF, so that the point A may be on D, and the straight line AB upon DE, the point B shall coincide with the point E, because AB is equal to DE ; and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also the point C shall coincide with the point F, because the straight line AC is equal to DF: But the point B coincides with

the point E. Therefore the base BC shall coincide with the a 10. Ax. base EF a, and be equal to it b. Wherefore the whole triangle

ABC shall coincide with the whole triangle DEF, and be equal b 8. Ax. to it b; and the other angles of the one shall coincide with the

remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those fides equal to one another, their bases shall likewise be equal, and the triangles be equal, and their other angles to which the equal fides are opposite, shall be equal, each to each. Which was to be demonstrated.

PROP. V. THEOR.
HE angles at the base of an Isosceles triangle are

equal to one another; and, if the equal fides be produced, the angles upon the other side of the base fhall be equal.

Let ABC be an Ifosceles triangle, of which the fide AB is equal to AC, and let the straight lines AB, AC be produced to D and E, the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE.

In BD, take any point F, and from AE, the greater, cut off a 3. 1. AG equal a to AF, the less, and join FC, GB.

Because

Because AF is equal to AG, and AB to AC, the two fides Book I. FA, AC are equal to the two GA, AB, each to each ; and

; they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is equal b to the base GB, and

b 4. 1. the triangle AFC to the triangle AGB ; and the remaining angles of the one are equal b to the remaining angles of the

B other, each to each, to which the equal fides are opposite, viz. the angle ACF to the angle ABG, and the angle AFC to the ky

G angle AGB: And because the whole AF is

D equal to the whole AG, of which the parts AB, AC are equal; the remainder BF shall be equal c to the

C 3. AZ remainder CG; and FC was proved to be equal to GB; therefore the two sides BF, FC are equal to the two CG, GB, each to each; and the angle BFC is equal to the angle CGB, and the base BC is common to the two triangles BFC, CGB; wherefore the triangles are equal b, and their remaining angles, each to each, to which the equal fides are opposite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG: "And since it has been demonstrated, that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are also equal; the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC: And it has also been proved, that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore the angles at the bafe, &c. Q. E. D.

COROLLARY. Hence every equilateral triangle is also equi. angular.

PROP. VI. THEOR.
F two angles of a triangle be equal to one another,

the sides also which fubtend, or are opposite to, the equal angles, shall be equal to one another.

Let ABC be a triangle having the angle ABG equal to the angle ACB; the side AB is equal to the side AC.

For if AB be not equal to AC, one of them A is greater than the other: Let AB be the greater, and from it cut a off DB equal to D AC, the less, and join DC; therefore, because

23. I, in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB, each to each ; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is equal B

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